Environmental Engineering Reference
In-Depth Information
These are not particularly elegant formulae. Not surprisingly, unlike Galilean
relativity, the components of an acceleration are not the same in all inertial
frames. In particular, a constant acceleration in one inertial frame is not a constant
acceleration in a different inertial frame.
We can however define an acceleration four-vector as simply the rate of change
of the velocity four-vector with respect to the proper time, i.e.
d
V
d
τ
.
A
≡
(14.4)
Starting from Eq. (12.7) we can write
A
in terms of its components in some inertial
frame, i.e.
A
=
γ(
γc,
˙
γ
u
˙
+
γ
a
),
(14.5)
where the dot indicates differentiation with respect to time as measured in the
inertial frame and
a
=
u
. The length of this four-vector will turn out to be of some
use to us and the quickest way to figure it out is to compute it in the inertial frame
in which
u
0
. It does not matter that this frame is useful only for an instant in
time (after which the particle may have developed a non-zero velocity); an instant
in time is long enough. In this instantaneous rest frame,
A
=
=
(
0
,
a
)
and hence
a
2
.
A
·
A
=−
(14.6)
Since
A
is a four-vector, this result is valid in all other frames. Note that
a
is the
magnitude of the three-acceleration in the inertial frame in which the particle is
instantaneously at rest: it is often called the 'proper acceleration' of the particle.
14.1.1
Twins paradox
As an example, let us consider the so-called twins paradox. Suppose that one
twin accelerates away from the Earth at a constant proper acceleration equal to
g
,
leaving the other twin behind. This rate of acceleration will lead the astronaut twin
to feel their weight inside the spaceship. After 10 years they switch the rockets
on their spaceship such that for the next 10 years they decelerate also at a rate
g
. At which time they again reverse the rockets and accelerate (again at
g
) back
towards the Earth for 10 further years before finally reversing the rockets one last
time for their arrival back on Earth some 10 years later. According to the twin who
travelled in the spaceship they were absent for a total of 40 years. The question
is, how much time has elapsed on Earth between the departure and return of the
astronaut twin?
Let us consider the first 10 years of the journey. When the astronaut is travelling
at speed
u
relative to an observer on Earth we imagine that they are instantaneously
at rest in an inertial frame
S
moving at speed
u
relative to Earth. We know that
the acceleration is constant and that
v
=
0in
S
. Hence, using Eq. (14.2) gives
d
(γ (u)u)
d
t
γ(u)
3
a
g
=
=
(14.7)
